Step 1: Calculate the maximum DC current through the

inductor, I

the top of the chart and show that I

buck configuration. Thus, I

Step 2: Calculate the inductor Volts-sec product, E-T

according to the equations given from the chart. For the

Buck:

E-T

=(15 − 5) (5/15) (1000/50)

= 66V-µs.

with the oscillator frequency, f

Step 3: Using the graph with axis labeled “Discontinuous At

%I

maximum inductor current, I

discontinuity percentage.

In this example, the point of interest is where the 0.35A line

intersects with the 20% line. This is nearly the midpoint of the

horizontal axis.

Step 4: This last step is merely the translation of the point

found in Step 3 to the graph directly below it. This is accom-

plished by moving straight down the page to the point which

intercepts the desired E-T

66V-µs and the desired inductor value is 470 µH. Since this

example was for 20% discontinuity, the bottom chart could

have been used directly, as noted in step 3 of the chart

instructions.

For a full line of standard inductor values, contact Pulse

Engineering (San Diego, Calif.) regarding their PE526XX

series, or A. I. E. Magnetics (Nashville, Tenn.).

A more precise inductance value may be calculated for the

Buck, Boost and Inverting Regulators as follows:

BUCK

L= V

∆I

BOOST

L= V

∆I

INVERT

L= V

∆I

where

∆I

∆I

usually chosen based on the minimum load current expected

of the circuit. For the buck regulator, since the inductor

current I

∆I

∆I

∆I

∆I

where the Discontinuity Factor is the ratio of the minimum

load current to the maximum load current. For this example,

the Discontinuity Factor is 0.2.

The remainder of the components of Figure 15 are chosen

as follows:

C1 is the timing capacitor found in Figure 1.

C2

≥ V

where V

C3 is necessary for continuous operation and is generally in

the 10 pF to 30 pF range.

D1 should be a Schottky type diode, such as the 1N5818 or

1N5819.

BUCK WITH BOOSTED OUTPUT CURRENT

For applications requiring a large output current, an external

transistor may be used as shown in Figure 17. This circuit

steps a 15V supply down to 5V with 1.5A of output current.

The output ripple is 50 mV, with an efficiency of 80%, a load

regulation of 40 mV (150 mA to 1.5A), and a line regulation

of 20 mV (12V

≤ V

in

≤ 18V).

Component values are selected as outlined for the buck

regulator with a discontinuity factor of 10%, with the addition

of R4 and R5:

R4 = 10V

R5=(V

where:

V

V

tor.

V is the current limit sense voltage.

B

Figure 17 ).

I

I

∆I

L

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